Chapter 9.5, Problem 49PPSE

Solution

a.

To explain the error made by the student.

Given information:

The sum of the infinite geometric series is twice its first term.

A student says the common ratio of the series is $\frac{3}{2}$ .

Calculation:

Given that the geometric series is infinite and its sum is twice the first term of the series.

Since the series is infinite geometric series and it has a sum, the common ratio must satisfy the inequality $\left|r\right|<1$ .

The value $\frac{3}{2}$ is greater than one. So, the series will not converge.

Therefore, the common ratio of the series cannot be greater than 1.

b.

To find the common ratio of the series.

  $r=\frac{1}{2}$

Given information:

The sum of the infinite geometric series is twice its first term.

Calculation:

Substitute the values $S=2a_1$ in the formula $S=\frac{a_1}{1-r}$ and solve for $r$ .

  $\begin{array}{c}2a_1=\frac{a_1}{1-r}\\ 2=\frac{1}{1-r}\\ 1-r=\frac{1}{2}\\ r=1-\frac{1}{2}\\ r=\frac{1}{2}\end{array}$

Therefore, the common ratio of the series is $r=\frac{1}{2}$ .